[tex]\bf \stackrel{f(x)}{y}=|x^2-1|\implies y=\sqrt{(x^2-1)^2}\implies y=[(x^2-1)^2]^{\frac{1}{2}}
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\cfrac{dy}{dx}=\stackrel{chain~rule}{\cfrac{1}{2}[(x^2-1)^2]^{-\frac{1}{2}}\cdot 2x}\implies \cfrac{dy}{dx}=x[(x^2-1)^2]^{-\frac{1}{2}}
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\cfrac{dy}{dx}=\cfrac{x}{[(x^2-1)^2]^{\frac{1}{2}}}\implies \cfrac{dy}{dx}=\cfrac{x}{\sqrt{(x^2-1)^2}}
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\left. \cfrac{dy}{dx}=\cfrac{x}{|x^2-1|} \right|_{x=1}\implies \cfrac{1}{|1-1|}\implies \cfrac{1}{0}\implies und efined[/tex]
meaning, the function is not differentiable at x = 1, graph wise, it simply means is not a "smooth curve", instead is an abrupt edge, or a "cusp" or a "spike", so the graph may be continuous at x = 1, however, is the extremum is not a smooth curve, notice the picture below.
